Today s Outline. CS 362, Lecture 4. Annihilator Method. Lookup Table. Annihilators for recurrences with non-homogeneous terms Transformations

Transcription

1 Today s Outline CS 362, Lecture 4 Jared Saia University of New Mexico Annihilators for recurrences with non-homogeneous terms Transformations 1 Annihilator Method Lookup Table Write down the annihilator for the recurrence Factor the annihilator Look up the factored annihilator in the Lookup Table to get general solution Solve for constants of the general solution by using initial conditions (L a 0 ) b 0(L a 1 ) b 1... (L a k ) b k annihilates only sequences of the form: p 1 (n)a n 0 + p 2(n)a n p k(n)a n k where p i (n) is a polynomial of degree b i 1 (and a i a j, when i j) 2 3

2 Examples Example Consider the recurrence T (n) = 7T (n 1) 16T (n 2)+12T (n 3), T (0) = 1, T (1) = 5, T (2) = 17 Q: What does (L 3)(L 2)(L 1) annihilate? A: c 0 1 n + c 1 2 n + c 2 3 n Q: What does (L 3) 2 (L 2)(L 1) annihilate? A: c 0 1 n + c 1 2 n + (c 2 n + c 3 )3 n Q: What does (L 1) 4 annihilate? A: (c 0 n 3 + c 1 n 2 + c 2 n + c 3 )1 n Q: What does (L 1) 3 (L 2) 2 annihilate? A: (c 0 n 2 + c 1 n + c 2 )1 n + (c 3 n + c 4 )2 n Write down the annihilator: From the definition of the sequence, we can see that L 3 T 7L 2 T + 16LT 12T = 0, so the annihilator is L 3 7L L 12 Factor the annihilator: We can factor by hand or using a computer program to get L 3 7L 2 +16L 12 = (L 2) 2 (L 3) Look up to get general solution: The annihilator (L 2) 2 (L 3) annihilates sequences of the form (c 0 n + c 1 )2 n + c 2 3 n Solve for constants: T (0) = 1 = c 1 + c 2, T (1) = 5 = 2c 0 + 2c 1 + 3c 2, T (2) = 17 = 8c 0 + 4c 1 + 9c 2. We ve got three equations and three unknowns. Solving by hand, we get that c 0 = 1,c 1 = 0,c 2 = 1. Thus: T (n) = n2 n + 3 n 4 5 Example (II) At Home Exercise Consider the recurrence T (n) = 2T (n 1) T (n 2), T (0) = 0, T (1) = 1 Write down the annihilator: From the definition of the sequence, we can see that L 2 T 2LT +T = 0, so the annihilator is L 2 2L + 1 Factor the annihilator: We can factor by hand or using the quadratic formula to get L 2 2L + 1 = (L 1) 2 Look up to get general solution: The annihilator (L 1) 2 annihilates sequences of the form (c 0 n + c 1 )1 n Solve for constants: T (0) = 0 = c 1, T (1) = 1 = c 0 + c 1, We ve got two equations and two unknowns. Solving by hand, we get that c 0 = 0,c 1 = 1. Thus: T (n) = n Consider the recurrence T (n) = 6T (n 1) 9T (n 2), T (0) = 1, T (1) = 6 Q1: What is the annihilator of this sequence? Q2: What is the factored version of the annihilator? Q3: What is the general solution for the recurrence? Q4: What are the constants in this general solution? (Note: You can check that your general solution works for T (2)) 6 7

3 Non-homogeneous terms Example Consider a recurrence of the form T (n) = T (n 1) + T (n 2) + k where k is some constant The terms in the equation involving T (i.e. T (n 1) and T (n 2)) are called the homogeneous terms The other terms (i.e.k) are called the non-homogeneous terms In a height-balanced tree, the height of two subtrees of any node differ by at most one Let T (n) be the smallest number of nodes needed to obtain a height balanced binary tree of height n Q: What is a recurrence for T (n)? A: Divide this into smaller subproblems To get a height-balanced tree of height n with the smallest number of nodes, need one subtree of height n 1, and one of height n 2, plus a root node Thus T (n) = T (n 1) + T (n 2) Example Example Let s solve this recurrence: T (n) = T (n 1) + T (n 2) + 1 (Let T n = T (n), and T = T n ) We know that (L 2 L 1) annihilates the homogeneous terms Let s apply it to the entire equation: (L 2 L 1) T n = L 2 T n L T n 1 T n = T n+2 T n+1 T n = T n+2 T n+1 T n = 1, 1, 1, This is close to what we want but we still need to annihilate 1, 1, 1, It s easy to see that L 1 annihilates 1, 1, 1, Thus (L 2 L 1)(L 1) annihilates T (n) = T (n 1) + T (n 2) + 1 When we factor, we get (L φ)(l ˆφ)(L 1), where φ = and ˆφ =

4 Lookup General Rule Looking up (L φ)(l ˆφ)(L 1) in the table We get T (n) = c 1 φ n + c 2ˆφ n + c 3 1 n If we plug in the appropriate initial conditions, we can solve for these three constants We ll need to get equations for T (2) in addition to T (0) and T (1) To find the annihilator for recurrences with non-homogeneous terms, do the following: Find the annihilator a 1 for the homogeneous part Find the annihilator a 2 for the non-homogeneous part The annihilator for the whole recurrence is then a 1 a Consider T (n) = T (n 1) + T (n 2) + 2. The residue is 2, 2, 2, and The annihilator is still (L 2 L 1)(L 1), but the equation for T (2) changes! Consider T (n) = T (n 1) + T (n 2) + 2 n. The residue is 1, 2, 4, 8, and The annihilator is now (L 2 L 1)(L 2)

5 Consider T (n) = T (n 1) + T (n 2) + n. The residue is 1, 2, 3, 4, The annihilator is now (L 2 L 1)(L 1) 2. Consider T (n) = T (n 1) + T (n 2) + n 2. The residue is 1, 4, 9, 16, and The annihilator is (L 2 L 1)(L 1) In Class Exercise Consider T (n) = T (n 1) + T (n 2) + n 2 2 n. The residue is 1 1, 4 4, 9 8, 16 16, and the The annihilator is (L 2 L 1)(L 1) 3 (L 2). Consider T (n) = 3 T (n 1) + 3 n Q1: What is the homogeneous part, and what annihilates it? Q2: What is the non-homogeneous part, and what annihilates it? Q3: What is the annihilator of T (n), and what is the general form of the recurrence? 18 19

6 Limitations Transformations Idea Our method does not work on T (n) = T (n 1)+ n 1 or T (n) = T (n 1) + lg n The problem is that n 1 and lg n do not have annihilators. Our tool, as it stands, is limited. Key idea for strengthening it is transformations Consider the recurrence giving the run time of mergesort T (n) = 2T (n/2) + kn (for some constant k), T (1) = 1 How do we solve this? We have no technique for annihilating terms like T (n/2) However, we can transform the recurrence into one with which we can work Transformation Now Solve Let n = 2 i and rewrite T (n): T (2 0 ) = 1 and T (2 i ) = 2T ( 2i 2 ) + k2i = 2T (2 i 1 ) + k2 i Now define a new sequence t as follows: t(i) = T (2 i ) Then t(0) = 1, t(i) = 2t(i 1) + k2 i We ve got a new recurrence: t(0) = 1, t(i) = 2t(i 1) + k2 i We can easily find the annihilator for this recurrence (L 2) annihilates the homogeneous part, (L 2) annihilates the non-homogeneous part, So (L 2)(L 2) annihilates t(i) Thus t(i) = (c 1 i + c 2 )2 i 22 23

7 Reverse Transformation Success! We ve got a solution for t(i) and we want to transform this into a solution for T (n) Recall that t(i) = T (2 i ) and 2 i = n t(i) = (c 1 i + c 2 )2 i (1) T (2 i ) = (c 1 i + c 2 )2 i (2) T (n) = (c 1 lg n + c 2 )n (3) = c 1 n lg n + c 2 n (4) = Θ(n lg n) (5) Let s recap what just happened: We could not find the annihilator of T (n) so: We did a transformation to a recurrence we could solve, t(i) (we let n = 2 i and t(i) = T (2 i )) We found the annihilator for t(i), and solved the recurrence for t(i) We reverse transformed the solution for t(i) back to a solution for T (n) Now Solve Consider the recurrence T (n) = 9T ( n 3 ) + kn, where T (1) = 1 and k is some constant Let n = 3 i and rewrite T (n): T (2 0 ) = 1 and T (3 i ) = 9T (3 i 1 ) + k3 i Now define a sequence t as follows t(i) = T (3 i ) Then t(0) = 1, t(i) = 9t(i 1) + k3 i t(0) = 1, t(i) = 9t(i 1) + k3 i This is annihilated by (L 9)(L 3) So t(i) is of the form t(i) = c 1 9 i + c 2 3 i 26 27

8 Reverse Transformation In Class Exercise t(i) = c 1 9 i + c 2 3 i Recall: t(i) = T (3 i ) and 3 i = n t(i) = c 1 9 i + c 2 3 i T (3 i ) = c 1 9 i + c 2 3 i T (n) = c 1 (3 i ) 2 + c 2 3 i = c 1 n 2 + c 2 n = Θ(n 2 ) Consider the recurrence T (n) = 2T (n/4) + kn, where T (1) = 1, and k is some constant Q1: What is the transformed recurrence t(i)? How do we rewrite n and T (n) to get this sequence? Q2: What is the annihilator of t(i)? What is the solution for the recurrence t(i)? Q3: What is the solution for T (n)? (i.e. do the reverse transformation) A Final Example A Final Example Not always obvious what sort of transformation to do: Consider T (n) = 2T ( n) + log n Let n = 2 i and rewrite T (n): T (2 i ) = 2T (2 i/2 ) + i Define t(i) = T (2 i ): t(i) = 2t(i/2) + i This final recurrence is something we know how to solve! t(i) = O(i log i) The reverse transform gives: t(i) = O(i log i) (6) T (2 i ) = O(i log i) (7) T (n) = O(log n log log n) (8) 30 31

9 Todo HW 1 Start Chapter 15 in text 32